4 votes
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Why is this differential injective? Lee Smooth Manifolds Proposition 5.3

This is true in general: If $f\colon X \to Y$ and $g\colon Y \to X$ are two functions satisfying $$ g\circ f = \operatorname{id}_X, $$ then $f$ is injective. If $x,y \in X$ are such that $f(x) = f(y)$,...
noam.szyfer's user avatar
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3 votes
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Confused by this proof in Jech's set theory

Turning some comments into an answer. Since $U\subseteq\mathcal N^2$ is $\mathbf{\Sigma}^0_\alpha$-universal and $A_n\subseteq\mathcal N$ is $\mathbf{\Pi}^0_\alpha$, so that $\mathcal N-A_n$ is $\...
Alessandro Codenotti's user avatar
2 votes

Why is it that: "If polynomials have integer coefficients, the roots of those polynomials will be a divisor (factor) of the constant term"?

The argument presented in the body of your question is perfectly sound [with a missing minus sign: a typo?] and is worth digesting. If you find that the notation is what's holding you back, then ...
Randall's user avatar
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2 votes
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Why we have to proof both $Q$ and $R$ in $P\implies (Q\lor R)$

In general, proving that $P\implies Q$ is sufficient to prove that $P\implies Q\lor R$. Sufficient, but not necessary, and in fact, often times, proving $P\implies Q$ may be impossible! Take, for ...
5xum's user avatar
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2 votes

Why is this differential injective? Lee Smooth Manifolds Proposition 5.3

Noam Szyfer's answer is fine, but here is another approach based on the fact that $\mathrm{d}\pi_M$ and $\mathrm{d}\gamma_x$ are linear maps. It is injective if and only if its kernel is trivial, i.e. ...
Abezhiko's user avatar
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1 vote
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Let $R$ be a UFD and $f(x)\in R[x]$ with $d$ as the content of the polynomial. Let $f(x)=df_1(x).$ Prove that $f_1(x)$ is a primitive polynomial.

By the properties of the gcd (valid in any gcd domain), $$\gcd(a_0,a_1,\ldots,a_n)=\gcd(cq_0,cq_1,\ldots cq_n)=c\cdot \gcd(q_0,q_1,\ldots q_n)=\\\gcd(a_0,a_1,\ldots,a_n)\cdot \gcd(q_0,q_1,\ldots q_n)$$...
Julio Puerta's user avatar
1 vote
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Logic behind contrapositive proofs that involves De Morgan's Laws

There are in fact several issues with your approach. Here are few observations: The statement you want to prove is one of arithmetic, and should rather be formalized in first-order logic (you need ...
Julio Di Egidio's user avatar
1 vote
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Change of coordinates in vector fields on Heisenberg group

You must use the chain rule, because the two sets of coordinates are functions of each other. For instance, the derivative with respect to $t = t(y_1,y_2,\tau)$ is given by $$ \frac{\partial}{\partial ...
Abezhiko's user avatar
  • 7,130
1 vote
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Doubt in proof of Prop 2.2.2 in Kesavan's Functional Analysis

Here's a way to select the sequence: From $\inf_{w\in W}\|x_{n_1}-x_{n_2}+w\|_V < \frac{1}{2}$ we get that there exists a $w_2 \in W$ such that $\|x_{n_1}-(x_{n_2}-w_2)\|_V < \frac{1}{2}$. Set $...
Giorgos Giapitzakis's user avatar
1 vote
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Understanding the proof of Theorem 10.1 in Montgomery & Vaughan's Multiplicative Number Theory

It means (I think) to consider the following path: That is, the rotated path is $\gamma(t) = \frac{t}{\sqrt{z}} = \frac{t}{ \sqrt{|z|}} e^{-i\frac{ arg z}{2}}$, so we also scale by $\frac{1}{\sqrt{|z|...
ploosu2's user avatar
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1 vote
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Let $A\subseteq \Bbb R^n$, then $(\overline{A^c})^c=\mathring A$

For the parts in blue: The statement $A^{{\complement}~\bigcirc}\subseteq A^\complement$ follows from the fact that $E^\bigcirc\subseteq E$ for any set $E$. The assertion $\bar A\subseteq A^{\...
K.defaoite's user avatar
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1 vote
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Use of intermediate value theorem to show that $p = (\cos 2\pi x, \sin 2 \pi x)$ is a covering map for $S^1$.

As Munkres writes, the fact that $p$ is a covering map comes from elementary properties of the sine and cosine functions. Which properties do we need? $\sin$ is strictly monotonically increasing on ...
Paul Frost's user avatar
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1 vote
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"Converse" to Chinese Remainder Theorem

First, this is only true because $q_1$ and $q_2$ are coprime $\iff (q_1,q_2)=1$: Suppose $\color{red}a \equiv \color{blue}{a_1}q_2+\color{green}{a_2}q_1\equiv \color{blue}{a_3}q_2+\color{green}{a_4}...
D S's user avatar
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